Divide the following complex numbers. $\dfrac{2-16i}{-3-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-3+i}$. $ \dfrac{2-16i}{-3-i} = \dfrac{2-16i}{-3-i} \cdot \dfrac{{-3+i}}{{-3+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(2-16i) \cdot (-3+i)} {(-3)^2 - (-i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(2-16i) \cdot (-3+i)} {(-3)^2 - (-i)^2} $ $ = \dfrac{(2-16i) \cdot (-3+i)} {9 + 1} $ $ = \dfrac{(2-16i) \cdot (-3+i)} {10} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({2-16i}) \cdot ({-3+i})} {10} $ $ = \dfrac{{2} \cdot {(-3)} + {-16} \cdot {(-3) i} + {2} \cdot {1 i} + {-16} \cdot {1 i^2}} {10} $ $ = \dfrac{-6 + 48i + 2i - 16 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-6 + 48i + 2i + 16} {10} = \dfrac{10 + 50i} {10} = 1+5i $